# Expected outcome

Three riflemen A, B, and C take turns shooting at a target. The first rifleman to hit the target gets 2002 dollars. A shoots first, B second, and C third, after which the cycle repeats again with A, until one of the riflemen hits the target. Each hits the target with probability 0.5. What is rifleman A's expected winnings in dollars?

According to me, the events are dependent on each other. So, if A hits the target then B and C do not have a chance to get the money. It is the same logic for B and C. We do not know how many times the cycle is repeated until someone wins. So for A to hit the target in the n-th cycle the probability of getting 2002 dollars is (1/2)^n. The three of them have equal probability of getting the money or not. The thing that bothers me is that we do not how many times each of them tries to hit the target until he wins. Does it have anything to do with the answer? Or independently of the number of cycles we can find it?

• Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Sep 12 '17 at 8:03
• Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 5xum Sep 12 '17 at 8:03
• I agree with 5xum – neonpokharkar Sep 12 '17 at 8:05

## 1 Answer

I'll give you a hint:

Let $2002 = w$, for wins.

You want to calculate $$w\sum_{t=0}^{\infty}P\{X=3t+1\}$$ where $X$ is the random variable that is the round that someone hits the target.

$$P\{X=1\}=0.5\\P\{X=2\}=0.25\\...\\P\{X=n\}=2^{-n}$$

because the shooters hit their mark around half the time. You can see this translates into a simple geometric series which (I hope) you can do yourself.